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## Main Question or Discussion Point

I'm having problems showing that Newton's second law of motion stays invariant (has the same form) under a Galilean transformation. If we write the general Galilean transformation as

[itex]t=t'+t_{t}[/itex]

[itex]\bar{x}=R\bar{x}'+\bar{u}t'+\bar{t}_{\bar{x}}[/itex]

where R an orthogonal transformation, then velocity and acceleration transform like this

[itex]\bar{v}(t)=\frac{d\bar{x}}{dt'}\frac{dt'}{dt} =R\bar{v}'(t')+\bar{u}[/itex]

[itex]\bar{a}(t)=\frac{d\bar{v}}{dt'}\frac{dt'}{dt} =R\bar{a}'(t')[/itex]

Therefore we can write for Newton's second law of motion

[itex]\bar{F}=m\bar{a}(t)=mR\bar{a}'(t')[/itex]

If this laws is invariant under a Galilean transformation then

[itex]\bar{F}'=m\bar{a}'(t')[/itex]

In other words, proving that Newton's second law of motion is invariant under a Galilean transformation, reduces to proving that

[itex]\bar{F}=R\bar{F}'[/itex]

However, I do not see why this should be the case. Can anyone shed some light on this?

[itex]t=t'+t_{t}[/itex]

[itex]\bar{x}=R\bar{x}'+\bar{u}t'+\bar{t}_{\bar{x}}[/itex]

where R an orthogonal transformation, then velocity and acceleration transform like this

[itex]\bar{v}(t)=\frac{d\bar{x}}{dt'}\frac{dt'}{dt} =R\bar{v}'(t')+\bar{u}[/itex]

[itex]\bar{a}(t)=\frac{d\bar{v}}{dt'}\frac{dt'}{dt} =R\bar{a}'(t')[/itex]

Therefore we can write for Newton's second law of motion

[itex]\bar{F}=m\bar{a}(t)=mR\bar{a}'(t')[/itex]

If this laws is invariant under a Galilean transformation then

[itex]\bar{F}'=m\bar{a}'(t')[/itex]

In other words, proving that Newton's second law of motion is invariant under a Galilean transformation, reduces to proving that

[itex]\bar{F}=R\bar{F}'[/itex]

However, I do not see why this should be the case. Can anyone shed some light on this?